Supermarket cashiers

When at the supermarket are opened only 2 cash people waiting in the front approximately 12 minutes. How many will shorten the average waiting time in a front where supermarket open another three cashiers?

Result

x =  7.2 min

Solution:

2:(2+3)=d:12  d=12⋅ 22+3=245=4.8 min  x=12−d=12−4.8=365=7.2=7.2  min 2:(2+3) = d:12 \ \\ \ \\ d = 12 \cdot \ \dfrac{ 2 }{ 2+3 } = \dfrac{ 24 }{ 5 } = 4.8 \ min \ \\ \ \\ x = 12-d = 12-4.8 = \dfrac{ 36 }{ 5 } = 7.2 = 7.2 \ \text{ min }

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